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Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

Let $|f(z)|\le K$ on $\mathrm{Re}z>0$. If $\mathrm{Re}z>c$, the circle $|w-z|=c$ lies in that half-plane. The Cauchy derivative formula givesThe line segment joining $z_1$ to $z_2$ remains in $\mathrm{Re}z>c$, soThus one may take $M=K/c$, as in the lipschitz bound inside a bounded analytic half-plane.

If (i) holds, the submodule theorem for free modules over a principal ideal domain says that any submodule $N\subseteq R^n$ is free. Its rank is at most $n$, since tensoring the inclusion with $F$ embeds $N{\otimes }_{R}F$ into $F^n$. Hence $N$ has at most $n$ generators, proving (iv).

Conversely, apply (iv) with $n=1$. Every ideal is an $R$-submodule of $R$ and therefore has one generator, which is (i). Thus all four conditions are equivalent.

Assume (iii). Because $R$ is a Noetherian ring, every ideal $I\subseteq R$ is finitely generated. Regard $I$ as an $R$-submodule of $F$. By (iii), $I=Rx$ for some $x\in F$; since $x\in I\subseteq R$ unless $I=0$, this is a principal ideal of $R$. Thus every ideal is principal and (i) holds.

Together with the previous two implications, this proves (i), (ii), and (iii) equivalent.

Assume (ii), and let an $R$-submodule $M\subseteq F$ be generated by $x_1,\dots ,x_m$. Choose a common nonzero denominator $d$ and write $x_j=a_j/d$ with $a_j\in R$. Repeated application of (ii) shows thatfor one element $g$: the gcd at each step is a linear combination of the elements processed so far. Thereforeso (iii) holds. This is the finite-ideal property defining a Bézout domain.

Assume (i). For $a,b\in R$, write the ideal $(a,b)=(d)$. Then $d=ra+sb$ for some $r,s\in R$. Since $a,b\in (d)$, the element $d$ divides both; and every common divisor of $a,b$ divides the displayed linear combination $d$. Thus $d$ is a greatest common divisor and (ii) holds.

True. Suppose $A,B$ are real and $AP=PB$ for an invertible complex matrix $P=X+iY$, with $X,Y$ real. Taking real and imaginary parts givesThe polynomial $q(t)=\det (X+tY)$ is not identically zero, since $q(i)=\det P\ne 0$. Choose a real $t$ outside its finite zero set. Then $Q=X+tY$ is real and invertible, and $AQ=QB$. Thus $A$ and $B$ are conjugate over $\mathbb{R}$, as stated by complex similarity of real matrices implies real similarity.

False. The real matrixhas the two distinct complex eigenvalues $i$ and $-i$, so it is diagonalizable over $\mathbb{C}$. It has no real eigenvalue and hence no real eigenbasis, so it is not diagonalizable over $\mathbb{R}$.

True. Since $\alpha$ is invertible, every eigenvalue $\lambda$ is nonzero. If $\alpha$ had a nontrivial Jordan block $\lambda I+N$, then for the stated positive integer $N_0$,would still have a nonzero nilpotent part and could not be diagonalizable. Therefore all Jordan blocks of $\alpha$ have size one. This also proves diagonalizability inherited from an invertible power.

False. At eigenvalue $2$, algebraic multiplicity two and minimal-polynomial exponent two force one block $J_2(2)$. At eigenvalue $1$, the blocks have total size five and largest size two, so there are only two possible partitions,Hence there are exactly two conjugacy classes satisfying the given polynomial data, represented byandThere cannot be three mutually non-conjugate examples.

True. By Jordan normal form, the characteristic polynomial records the total size of the blocks for each eigenvalue, while the minimal polynomial records the largest block for each eigenvalue. In dimension three this determines every block partition: for algebraic multiplicity three the possibilities $3$, $2+1$, and $1+1+1$ have largest block sizes three, two, and one; multiplicities one and two are equally immediate. Thus the two matrices have the same Jordan form and are conjugate. This is the characteristic and minimal polynomials determine similarity in dimension three phenomenon.

The null hypothesis is that mortality is independent of carbolic-acid use, equivalently that the two mortality probabilities are equal. The two-sided alternative is that they differ.

Use the Chi-squared test of independence, equivalently the likelihood-ratio statistic, with Pearson's statisticwhere $E_{ij}$ are the fitted counts under independence using the observed row and column totals. Under the null, $X^2$ is asymptotically ${\chi }_1^2$. A size-$\alpha$ test rejects when

For a null parameter space ${\Theta }_0\subseteq \Theta$, the generalized likelihood ratio isThe Wilks theorem states, under standard regularity conditions and under the null, that

If $A$ has full column rank, then for every nonzero $h$,Thus $A^{T}A$ is positive definite and invertible, and the normal equation has the unique solution